After passing through a gold film of thickness $5 \times 10^{-3} \ m$,the intensity of an $X$-ray beam decreases to $36.8\%$ of its initial intensity. The absorption coefficient is ....... $m^{-1}$.

The energy of an electromagnetic radiation is $14.4 \, KeV$. To which part of the electromagnetic spectrum does it belong?

$\Delta \lambda$ is the difference between the wavelength of the $K_\alpha$ line and the minimum wavelength of the continuous $X$-ray spectrum when the $X$-ray tube is operated at a voltage $V$. If the operating voltage is changed to $V / 3$,then the above difference is $\Delta \lambda^{\prime}$. Then:

$X$-ray beam of intensity $I_0$ passes through an absorption plate of thickness $d$. If the absorption coefficient of the material of the plate is $\mu$,the correct statement regarding the transmitted intensity $I$ of the $X$-ray is:

The minimum wavelength of $X$-rays produced by an electron accelerated through a potential difference of $V$ volts is proportional to:

An $X$-ray has a wavelength of $0.010 \mathring{A}$. Its momentum is